Is ${313038}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {313038}= &&{3}\cdot100000+ \\&&{1}\cdot10000+ \\&&{3}\cdot1000+ \\&&{0}\cdot100+ \\&&{3}\cdot10+ \\&&{8}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {313038}= &&{3}(99999+1)+ \\&&{1}(9999+1)+ \\&&{3}(999+1)+ \\&&{0}(99+1)+ \\&&{3}(9+1)+ \\&&{8} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {313038}= &&\gray{3\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {3}+{1}+{3}+{0}+{3}+{8} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${313038}$ is divisible by $9$ if ${ 3}+{1}+{3}+{0}+{3}+{8}$ is divisible by $9$ Add the digits of ${313038}$ $ {3}+{1}+{3}+{0}+{3}+{8} = {18} $ If ${18}$ is divisible by $9$ , then ${313038}$ must also be divisible by $9$ ${18}$ is divisible by $9$, therefore ${313038}$ must also be divisible by $9$.